# If $f:\mathbb{R}^n \to \mathbb{R}^n$ is continuous with convex image, and locally 1-1, must it be globally 1-1?

For $f:\mathbb{R}\to \mathbb{R}$ which is continuous, being locally 1-1 implies being globally 1-1, see here. This is not true for a general mapping $f:\mathbb{R}^n\to \mathbb{R}^n$. My intuition as to the source of this incongruity is that while continuity preserves connectedness, it does not preserve convexity.

This illustration (from Buck's Advanced Calculus) of a locally injective, not globally injective function from $\mathbb{R}^2$ to $\mathbb{R}^2$ somewhat supports my intuition...note the image is not convex. If it were convex, it would seem there would need to be some "critical point" (not assuming the existence of the derivative) where $f$ was not locally 1-1.

Is there a theorem that if $f: S \to \mathbb{R}^n$ is continuous, for $S\subset \mathbb{R}^n$ convex, with convex image, locally 1-1, then it must be 1-1 on all of $S$?

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I think you need some kind of compactness condition, otherwise you could remove part of the boundary of the region in the figure and sort of fold it in half. – augurar Jul 15 '13 at 23:30
By "convex image" do you mean just that $f(\Bbb R^n)$ is convex, or that the image of each convex set is convex? – dfeuer Jul 15 '13 at 23:39
@dfeuer I mean that the image of the domain is convex (the domain may be any $S\subset \mathbb{R}^n$, $S$ convex). – Eric Auld Jul 15 '13 at 23:42
I added a link to your previous question, for easy reference. – Andrés Caicedo Jul 16 '13 at 3:00
I think it may be more promising to consider functions that preserve convexity (or perhaps simple connectedness), so that the image of every convex[/simply connected] set is convex[/simply connected]. – dfeuer Jul 16 '13 at 20:11

In $\Bbb R^2$, it's easy to "paint" a counter-example.

Just take a wide, thin paintbrush, dip it in paint, and use it to paint a very large disk. Keep your strokes very smooth, never twisting the brush in place and never making any sudden movements.

This gives you a locally one-to-one, continuous function from a long, thin rectangular region (the paint trail) to the disk. The paint trail and the disk are both convex and compact, but the function is not one-to-one.

Clarification: How can we be sure our strokes are smooth enough? Let one face of the paintbrush be considered the front, and the other the back. Move the brush in a sequence of arcs, choosing a pivot point along the line running from the left end of the paintbrush to the right end, but lying outside the brush. Move the paintbrush in a "forward" direction each time.

It will be easiest to paint an annulus around the edge first, as one arc, and then fill in the rest.

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One can modify the pictured example so as to have domain, say, the set of $(x,y) \in \mathbb{R}^2$ with $0 < x < 1$. Imagine the function being given by taking a line segment of width $1$, mapping the center of the line segment to $f(\frac{1}{2},y)$, and then carrying around the rest of the line segment linearly. Thus we need a continuous function $g: \mathbb{R} \rightarrow \mathbb{R}^2$ so as to set $f(\frac{1}{2},y) = g(y)$. If $g$ is smooth with nowhere vanishing derivative, $f$ is locally injective. It is easy to choose such a $g$ so as to make $f$ surjective -- e.g. make sure that every once in a while $g$ traces out the circle of radius $\frac{n}{2}$ centered at the origin for each $n \in \mathbb{Z}^+$. -- and not injective -- each make sure $g$ does what I said above at least twice for each positive integer $n$.

More precisely: I want the segment to be oriented so that for all $y \in \mathbb{R}$, it is perpendicular to the tangent vector $g'(y)$. Once I specify its orientation is at $y = 0$ (take it to be horizontal; it doesn't matter) that determines it for all $y$.

Since the domain of $f$ is diffeomorphic to $\mathbb{R}^2$, we can pull this back to get a (smooth) counterexample $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$.

Added: The above construction can also be used to give an $f$ such that for all $v \in \mathbb{R}^2$, $f^{-1}(v)$ is countably infinite. Countably infinite is the largest possible preimage for a locally injective function on $\mathbb{R}^2$ (or any $\sigma$-compact $T_1$ topological space): cover $\mathbb{R}^2$ by countably many compact subsets. If any preimage were uncountable, its intersection with some compact subset would be uncountable, and thus the preimage of $f^{-1}(v)$ would have an accumulation point, at which $f$ is not locally injective.

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Isn't that about what I said? – dfeuer Jul 16 '13 at 2:32
@dfeuer: yes, it's very close. The upvote on your answer is from me. Let me say that I spent the better part of an hour doodling in my notebook to answer this question -- and I didn't read your answer until I was finished -- so when I finally got something I decided to post it. – Pete L. Clark Jul 16 '13 at 2:49
@dfeuer: By the way, why no one else has upvoted your correct answer is a mystery to me. Unfortunately I think that this site is getting so many questions right now that the number of people who are reading answers to any given question (and especially, questions which require nontrivial background) is getting rather small. – Pete L. Clark Jul 16 '13 at 2:54
Ah, I've definitely spent too long writing answers and found them overtaken quite a few times. Thanks for the upvote. – dfeuer Jul 16 '13 at 3:01
@dfeuer: Well, I guess I need the space to be T1 as well. I have in mind the following straightforward argument: an accumulation point $x \in f^{-1}(v)$ lies (by definition) in the closure of $f^{-1}(v) \setminus \{x\}$. Since $f$ is continuous -- and $\{x\}$ is closed! -- this means $f(x) = v$. Then every neighborhood of $x$ has an element $y \neq x$ such that $f(y) =f(x)$. I hope I haven't missed some subtlety... – Pete L. Clark Jul 16 '13 at 4:27

There need not be any critical points. What I believe is the standard example Buck is giving there is $$f\colon\mathbb R^2\to\mathbb R^2, \quad f(x,y)=(e^x\cos y,e^x\sin y).$$ If you know complex variables, you'll recognize $f$ as $e^z$.

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I don't think any holomorphic function will give you a counterexample. – tomasz Jul 15 '13 at 23:45
@tomasz: Well, I was wrong, but it's probably just topological. If the image is convex, it can't have any nontrivial connected covering spaces. Perhaps we need a properness hypothesis. – Ted Shifrin Jul 16 '13 at 0:10